# S 16-5 - Math 200 Spring 2010 Worksheet#27 Name_K_°_‘...

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Unformatted text preview: Math 200, Spring 2010 Worksheet #27 Name _K_°_‘[_______ Section 16-5 Surface Integrals of Vector Fields 1. Evaluate the surface integral ”3 F . (18 for the vector ﬁeld F(x, y,z) = yi+ xj+ 22k and S is the helicoid with parametrization d)(u,v) = (u cos v,u sin v,v), O S u S 1, 0 S v S :r with upward orientation. ~—* Br} . e ’9 A 11“)“ "(COSV)SH~U,0) ._. 7:. __. t. ,3- K T _. }§_ n ‘-‘ luxTU :' (on! 5m: (5m, {0591.) v - av <“u5ml0 IA(¢J.SU I“) «an: LtCOSu l -- WHFL faint: ufiuan-J Since “~30 F(@(QU)) )=<U.S|.V\V)DLIGJU V") ESP 0L? 2 23:4 hag: o£f<u\$mu—ucos v+uu:;)lﬂaluatv :foJ(_ “CDSJV +hvzjdudu”Oj?da.uCOSJv1-auu ‘IZJIJ’COJJU-tlm O"y51h20+%V\$/fz=é1:;o 2. Evaluate the surface integral {IF - dS for the vector ﬁeld F(x, y,z )= xyi+4x2 j+ yzk and S IS the surface 2— — xey, 0 S x S 1, 0 S y S 1 , with upward orientation. " ‘UM‘an 2 2‘ Y :zxy)‘1’x1”z—) Pantomime: as agmfkog a“: ‘k ‘3‘ all”) 268 @(wﬁ Gyms") 05x5.) 0am 'ﬁ‘=(—-3:~gb:> F(—(x/7)): {X7} LIL"; W66 '1'? u? zi‘ggri'ez 17 um manna ma) . F-a's‘z g Ifl? whyf(ﬂx7€'ﬂ“lﬂ e7+yxe")dacd7 O 1 39:1 :3 _"'7 -.._ ‘51:]! d7“ fe’ayzx—e, o xeo 3. Evaluate the surface integral HF - dS for the vector ﬁeld F(x, y,z )= zi + xj +k and S is the boundary of the region enclosed by the upper hemisphere x2 + y2 + 22 =1, 2 2 O and the plane 2— - 0. No‘lte'. For a. closed Surface ‘Hvri is He loovwiar- o‘f 0- >0itel Team“ ‘91 +Le Convent-ton <5 Hart 4*? post-lure errenl‘dten is “W- 51“— ‘Fw whet. ‘HsQ “Emmi vectors Powd- oul‘waml ¥romb> “Thus 5, is He Upper Remustet-e cried-eel ofwmvlﬂ a»; \$1 :5 +Le bbﬁ'om ch59: onen’f-e-l downward to “'hk @(06)= “‘(SthCOSB 51M? Sine (05¢)Oilﬁ- P1059531? 5Evn(<9e)= SmCD(SmIPCOSB Smcpsm Cos<9> Fat/9w): (cosc9 <9m‘n9cose D _ 9",; Qt: ,70 J 0 SIFJS ffCOSQSMQCMB+5m¢COSB§mB+3detoStdd3-Jtp ofmwmsm offs: Tr cnszrsg‘ﬁ 4‘s“: 3‘ F “Rats: HF- +1045: g—MAhI'h—rﬁwo "yr-14.5: n- 1r: 0 4. Show that the ﬂux of F = e_, through a sphere centered at the origin does not depend on the radius of the sphere. r2 Qarqme’ma-ai-Ion 0'? a. sphere wrl-k f‘adtu: R C'ineae-l a‘l‘ +l~e aha?“ , Ql (,th (RsmfkwseﬂQSmcﬂmue, ﬂearcﬂlﬂj OSQfJTJGﬁCﬂir The Srhete, has OUl'uJaml Paladin-1 nbrmal W~ " R SmCP err on ~er Sphere eij PacltuJ R; F{§(¢e))_ - f2“:- ....5 __ ‘ K F'n - R1 fRSmCOf '-‘- \$mq9 Ear-gr :SMCp l’l‘ “ .. __, h, :1- T‘: 1? e “'1 gulf-3&5 " gﬂlpﬁlddlde -' I Sm 490107.49 : 211‘[-cosCP—] :9.“- e 6 a WP 596 +twtt ~H~e SVr‘Fﬁ-CP inleaml all 75’ does no‘l‘ depend am He mam R of- He slide-e. 5. Set up the integral (do not evaluate) that computes the ﬂux “5 F -dS for the given oriented surfaces: a. F(x, y, z) = (w— y, z, —x); S is the parallelogram with parameterization CD(u, v) = (u + 3v,v — 2u, 2v + 5), 0 S u,v S 1 with upwardgoihging normal. ~._)-—..a =._ _‘ .3 ___ L a K t“ _ {if-210) n : quTu 2 l—J. 05:: {‘H) 3 ‘7) an ofmr'i r 431,1) 3 ‘2 31°33;er Elf-3T0”: {ﬁle U 2U+5 u- 3U> '51“ng ‘15 I. nJA Bu-tHU-Hu—m “7:; an; = 4514-3104:) 203:;( lsh- -3Iv~— lo)aludu_ b. F(x,y,z)= ( 2c,2,y2 x2+ yz); Sis the hemisphere x2 +y2 +z2 =4, z>0 with outward pointing Farﬁﬁ‘rﬁ'l'nﬁ S. Efﬁe): '(QCMGSlnﬂ/IQSInGSmQ 3(0563) 0ﬁ\$4%J outward point inc; normal ‘ n - 2 quﬂer Jib G‘l'tr Fill) ((0/3)) IJLZLlrcos Q Sm chp L15th "“2951“? LLZZ'fécgses'ﬁcqsmes’Mp) Co: 61> DUX“ :' —-* .21 HF AS: JIF “an :fojﬁrf (coséﬁm (9+1!) SthaGSm‘lc'} +14 COS‘ﬂ 5‘" 3‘p)d90°(6 c. F(x, y, z) — (2,0,2 ); S has parameterization CD(u,v)=(v(e“1—e'"),v(e” — e’"),2v), 0 S u,v S 1, A 'Tllr normal. Awith upward pointing normal“;- M.‘ a: _. :1 ’q i K downwind \$61K Tu=<vfe_ -e)ve+e),0> n~ ulC'f)‘-‘(€+e o Gsuél, V ._ ‘— ——| Tl): (eh-e ,CM-e :2) 8+3“ 6”,?“ 3 Sauce. -r\ WWW) <2»; 0 w > — Hume“) Me 9'“) an 6. Compute the surface integral of F(x, y, z) = (x, y, z) for the following oriented surface: S is the boundary of the closed cylinder x2 + y2 S 1, 0 S 2 S 3 with normals pointing out of the cylinder. S = Sl U S2 U S3 , where S1 is the bottom, 2 = 0, x2 + y2 S 1 , with downward pointing normal, S2 is the top 2 = 3, x2 + y2 S1, with upward pointing normal, and S3 is the cylinder CD(t, z) = (cos(t), sin(t), z), 0 S t S 271', 0 s z 5 3 , with outward pointing normal. ILEdS: ”ﬁnds... I92F'ds+ILF'dS on 5.: 2919p) 2 (nose) rsiv‘te (3)0595311' oérﬁl F(§(Qr1):4rcose «5.5» T5: < ”“9 ”are 0) ,4 I y F-W=OS-_. _, Tr=(Cose,San§ o) 3‘ - IF'J. =0 '5‘ : L " k S 5‘ S 5. —rs-ne rcoseo : (@0er : u on “ ' \$1053 “ 5(61'):(Ptose)rgfhe 3) 0:086 5.5 0 “WWW. 2 2 ~95 2n airs-l {31.1.1.7li ‘5 ‘— It 3“- .— “hm-l £11386. F(@(6ryr( n " <02 °;?> UfW-W‘ foiva'e normal __‘ _J u u I *C656,r:,‘.,s 3c, »‘ Z 3 PAS F'n '5'- 3r D'%_d§_ ’ 2x 1 a I 5 5 ' [grdrdDCQ‘T- gr! =37? ‘-'- O+3n+6if °“S35§(e a) t 1 o o O i: " (559) SMQ '2) 0‘652’171 0&25? '3' : ((0:91 Since/o) : 911' I‘“ ”4055” 9M8 2;) (case m5 0) =1 SZIE-djz Iffdaa :2rr 3’ 'QT 7. Calculate the rate of ﬂow (volume/unit time) of a ﬂuid with velocity ﬁeld 5v(x, y,z)= (— y,x, 2) through the graph of z = 4 — x2 — y , 0 S x2 + y2 S 4 , with upward pointing normal. PGquEJWIiE a: Cs. Snark 0‘1: 01. coholtan 2 3309‘!) : 4-32.71. MW = Om W~><a71> ) 0 “3719+ r: = ("3’1“211‘5 45-) U (§(¥/73) ”' <"yJXJ th ‘7 > u :QLJ Z|ch:> “"Ml WV a n v r a ,7 n QXy+2x7++>< 7 =-'i’ X “L7 N19 \$5311.; _ ~---- ‘5 1‘ L 1 (-91%: 9w?) . ZKU'dS : fftlkbxv )0”! : T(LLl-rl)rdrde Homework #27 Reread Section 16.5 (Due 05/03) Do Section 16.5 Exercises: 5, 7, 9, 13, 14, 15, 17, 19, 21, 23, 26 Prepare for next class session: Read Section 17.1 ...
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